Use c = 3.00 × 10⁸ m/s, hc = 1240 eV nm and h = 4.14 × 10⁻¹⁵ eV s…

Step 1. The router. $\lambda = \dfrac{c}{f} = \dfrac{3.00 \times 10^{8}\ \text{m/s}}{5.0 \times 10^{9}\ \text{Hz}} = 0.060\ \text{m} = 6.00\ \text{cm}$.

Step 2. The thermal camera. $\lambda = 10.0\ \mu\text{m} = 10000\ \text{nm}$. So $E = \dfrac{1240\ \text{eV nm}}{10000\ \text{nm}} = 0.124\ \text{eV}$.

Step 3. The scanner. $\lambda = \dfrac{1240\ \text{eV nm}}{80000\ \text{eV}} = 0.0155\ \text{nm} = 15.5\ \text{pm}$.